Proofs

It is common to employ algebra and complex numbers to solve planar problems. Below, I'll reverse the action by interpreting the problem in geometric terms and solving the result as a problem in triangle geometry.

First note that the condition $a|bc|+b|ca|+c|ab|=0$ tells us that the angles between any two of the three vectors joining the origin to the points $A,B,C$ associated with complex numbers $a,b,c$ are all $120^{\circ}.$ Indeed, the lengths of the the three vectors $a|bc|,b|ca|,c|ab|$ are equal so that, since their sum is $0,$ they form an equilateral triangle.

i.e., (*). Quite obviously, the equality only holds for the equilateral triangle.

Leo Giugiuc has offered a shorter proof which was also suggested in the comments below. After establishing that the angles between the vectors $a,b,c$ (back to his original notations) are $120^{\circ},$ he observes that, say, $|a-b|$ is the third side of the triangle with sides $|a|$ and $|b|,$ opposite the $120^{\circ}$ angle. By the Cosine Law,

$|a-b|^{2}=|a|^{2}+|ab|+|b|^{2}\ge 3|ab|$

because $(|a|-|b|)^{2}=|a|^{2}-2|ab|+|b|^{2}.$ It follows that $|a-b|\ge \sqrt{3}\sqrt{|ab|}.$ Similar inequalities hold for $|b-c|$ and $|c-a|;$ the product of the three amounts to the required inequality.

Amit Itagi offered a variant:

When one od $a,b,c\,$ is $0\,$ the solution is trivial. Otherwise, let $a=|a|e^{i\alpha}$, $b=|b|e^{i\beta}$, $c=|c|e^{i\gamma}$. The constraint implies

$e^{i\alpha}+e^{i\beta}+e^{i\gamma}=0.$

Three unit complex numbers summing to $0$ means that the unsigned angle between any two of $\{a,b,c\}$ is $2\pi/3$. Thus, applying cosine rule to the triangle formed by $a$, $b$, and $a-b$,

Three Junior Problems from Vietnam $\left(\left(\begin{array}{ccc}0&y&z\\x&0&z\\x&y&0\end{array}\right)\left(\begin{array}\;a\\b\\c\end{array}\right)=\left(\begin{array}\;x\\y\\z\end{array}\right)\right)$

Four Integrals in One Inequality $\left(\displaystyle \small{\left(\int_a^bxf(x)dx\right)\left(\int_a^bf^2(x)dx\right)\left(\int_a^bx^3f(x)dx\right)\ge\frac{a^2b^2}{b-a}\left(\int_a^bf(x)dx\right)^4}\right)$